# Force of electromagnet on a wire [closed]

I'm trying to produce an equation for the force produced by an electromagnet on a wire. I'm very aware I don't know any vector calculus but I think as it's the special case (Perpendicular) I can just use Algebra.

The three equations I have used are:

\begin{align} I &= n_{e}Ave \tag{1} \\ F &= Bqv \tag{2} \\ B &= \mu_{o} n_{t}I \tag{3} \\ \end{align}

I then rearranged Equation $(1)$ putting it in terms of $v$ and substituted it into Equation $(2)$. I did the same for Equation $(3)$ subbing it into Equation $(2)$ to get:

$$F = (\mu_{o} n_{t}I)(e)\left( \dfrac {I} {n_{e}Ae}\right) = \dfrac {I^{2}\mu _{0}n_{t}} {n_{e}A} \, .$$

My First Question is to get the overall force, would you simply remove the number of electrons ($n_{e}$)?

And the more pressing concern, when I do dimensional analysis ( rearranged in terms of $\mu_{0}$ ) I am out by a factor of Length squared i.e the area of the wire ( in meter squared) should not be in the equation. Can I ask where I have gone wrong and if I need a fundamental rethink?

EDIT: For Clarification $n_{t}$ is the number of turns per unit length.

• The force of a constant magnetic field $B$ on a current $I$ flowing trough a wire of length $l$ is simply $F=Bl_{wire}I_{wire}$, which should give you the right units. The magnetic field of a solenoid is approximately $B=\mu_0 nI/l_{solenoid}$. Jun 29, 2015 at 22:02